MATH 409 Lecture 25

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End Exam 3 content

Lecture Slides

Course Evaluations.

Final exam this Friday, 15:00–17:00

Extra Office Hours:

• 4 Dec 12:30–14:00
• 5 Dec 12:00–14:00
• 6 Dec 10:00–12:00

Final Review

Topic List

Part 1: Axiomatic Model of the Real Numbers

• Axioms of an ordered field
• Completeness axiom
• Archimedean principle
• Mathematical Induction
• Binomial formula
• Countable and uncountable sets

Chapters 1.1–1.6, Appendix A

Part 2: Limits and Continuity

• Limits of Sequences
• Limit Theorems for Sequences
• Monotone sequences
• Bolzano-Weierstrass Theorem
• Cauchy Sequences
• Limits of Functions
• Limit theorems for functions
• Continuity of functions
• Extreme Value and Intermediate Value Theorems
• Uniform Continuity

Chapters 2.1–2.5, 3.1–3.4

Part 3a: Differential Calculus

• Derivative of a function
• Differentiability theorems
• Derivative of the inverse function
• Mean Value Theorem
• Taylor's formula
• l'Hôpital's rule

Chapters 4.1–4.5

Part 3b: Integral Calculus

• Darboux Sums, Riemann Sums, and Riemann integral
• Properties of integrals
• Fundamental Theorem of Calculus
• Integration by parts
• change of variable in an integral
• Improper Integrals, Absolute Integregrability

Chapters 5.1–5.4

Part 4: Infinite Series

• Convergence of series
• Comparison test and integral test
• Alternating series test (Leibniz Criterion)
• Absolute convergence
• Ratio test and Root test

Chapters 6.1–6.4

Theorems to Know

Archimedian Principle

For any real number $ϵ>0$, there exists a $n\in \mathbb{\mathbb{N}}$ such that $nϵ>1$.

Theorem. $\mathbb{\mathbb{Z}}$, $\mathbb{\mathbb{Q}}$, $\mathbb{\mathbb{N}}\times \mathbb{\mathbb{N}}$ are countable

Theorem. $\mathbb{ℝ}$ is uncountable.

Theorems on Limits

Squeeze Theorem. If $\underset{n\to \mathrm{\infty }}{\lim }{x}_n=\underset{n\to \mathrm{\infty }}{\lim }{y}_n=a$, and $x_n<w_n<y_n$ for all $n>N$, then $\underset{n\to \mathrm{\infty }}{\lim }{w}_n=a$.

Theorem. Any monotone sequence converges to a limit if bounded, and diverges to infinity otherwise.

Theorem. Any Cauchy sequence is convergent.

Theorems on Derivatives

Theorem. If $f$ and $g$ are differentiable at $a\in \mathbb{ℝ}$, then their sum $f+g$, difference $f-g$, and product $f\cdot g$ are also differentiable at $a$. Moreover,

$\begin{array}{rl}(f+g{)}^{\prime }(a)& =f^{\prime }(a)+g^{\prime }(a)\\ (f-g{)}^{\prime }(a)& =f^{\prime }(a)-g^{\prime }(a)\\ (f\cdot g{)}^{\prime }(a)& =f^{\prime }(a)g(a)+f(a)g^{\prime }(a)\\ \end{array}$

If, additionally, $g(a)\ne 0$, then the quotient $\frac{f}{g}$ is also differentiable at $a$ and

$\left(\frac{f}{g}\right)(a)=\frac{f^{\prime }(a)g(a)-f(a)g^{\prime }(a)}{(g(a){)}^2}$

Mean Value Theorem. If a function $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $c\in (a,b)$ such that $f(b)-f(a)=f^{\prime }(c)(b-a)$.

Theorems on Integrals

Theorem. [Linearity]. If $f$ and $g$ are integrable on $[a,b]$, then $f+g$ is also integrable on $[a,b]$ and

$\underset{a}{\overset{b}{\int }}(f(x)+g(x))\mathrm{d}x=\underset{a}{\overset{b}{\int }}f(x)\mathrm{d}x+\underset{a}{\overset{b}{\int }}g(x)\mathrm{d}x$

Theorem. If a function $f$ is integrable on $[a,b]$, then for any $c\in (a,b)$,

$\underset{a}{\overset{b}{\int }}f(x)\mathrm{d}x=\underset{a}{\overset{c}{\int }}f(x)\mathrm{d}x+\underset{c}{\overset{b}{\int }}f(x)\mathrm{d}x$

Theorems on Series

Integral Test. Suppose that $f:[1,\mathrm{\infty })\to \mathbb{ℝ}$ is positive and decreasing on $[1,\mathrm{\infty })$. Then the series $\sum _{n=1}^{\mathrm{\infty }}f(n)$ converges if and only if the function $f$ is improperly integrable on $[1,\mathrm{\infty })$.

Ratio Test. Let $\left\{a_n\right\}$ be a sequence of reals with $a_n\ne 0$ for large $n$. Suppose that a limit $r=\underset{n\to \mathrm{\infty }}{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}$ exists (or is finite), then

1. If $r<1$, then $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ converges absolutely.
2. If $r>1$, then $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ diverges.

Sample Problems

Problem 1 (20 pts)

Union of countable sets is also countable

Theorem. Suppose $E_1$, $E_2$, $E_3$, ... are countable sets. Prove that their union $E_1\cup E_2\cup E_3\cup \dots$ is also a countable set.

Proof. First we are going to show that $\mathbb{\mathbb{N}}\times \mathbb{\mathbb{N}}$ is countable.

Consider a relation $\prec$ on the set $\mathbb{\mathbb{N}}\times \mathbb{\mathbb{N}}$ such that $(n_1,n_2)\prec (m_1,m_2)$ if and only if either $n_1+n_2<m_1+m_2$ or else $n_1+n_2=m_1+m_2$ and $n_1<m_1$. (similar to a lexicographic order, but important difference). It is easy to see that $\prec$ is a strict linear order. Moreover, for any pair $(m_1,m_2)\in \mathbb{\mathbb{N}}\times \mathbb{\mathbb{N}}$, there are only finitely many pairs $(n_1,n_2)$ such that $(n_1,n_2)\prec (m_1,m_2)$. It follows that $\prec$ is a well-ordering.

Now we define inductively a mapping $F:\mathbb{\mathbb{N}}\to \mathbb{\mathbb{N}}\times \mathbb{\mathbb{N}}$ such that for any $n\in \mathbb{\mathbb{N}}$, the pair $F(n)$ is the least (relative to $\prec$) pair different from $F(k)$ for all natural numbers $k<n$. It follows from the construction that $F$ is bijective. The inverse mapping $F^{-1}$ can be given explicitly by

$F^{-1}(n_1,n_2)=\frac{1}{2}(n_1+n_2-2)(n_1+n_2-1)+n_1$

Thus $\mathbb{\mathbb{N}}\times \mathbb{\mathbb{N}}$ is a countable set.

Now suppose $E_1,E_2,\dots$ are countable sets. Then for any $n\in \mathbb{\mathbb{N}}$, there exists a bijective mapping $f_n:\mathbb{\mathbb{N}}\to E_n$. Let us define a map $g:\mathbb{\mathbb{N}}\times \mathbb{\mathbb{N}}\to E_1\cup E_2\cup \dots$ by $g(n_1,n_2)=f_{n_1}(n_2)$. Obviously $g$ is onto.

Since the set $\mathbb{\mathbb{N}}\times \mathbb{\mathbb{N}}$ is countable, there exists a sequence $p_1,p_2,p_3,\dots$ that forms a complete list of its elements. Then the sequence $g(p_1),g(p_2),g(p_3),\dots$ contains all elements of the union $E_1\cup E_2\cup E_3\cup \dots$. Although the latter sequence may include repetitions, we can choose a subsequence $\{g(p_{n_k})\}$ in which every element of the union appears exactly once. Note that the subsequence is infinite since each of the sets $E_1,E_2,\dots$ is infinite.

Now the map

$h:\mathbb{\mathbb{N}}\to E_1\cup E_2\cup E_3\cup \dots$

defined by

$h(k)=g(p_{n_k})$

for

$k=1,2,\dots$

is a bijection.

quod erat demonstrandum

Problem 2 (20 pts)

Evaluate following limits

Subproblem 2a

$\underset{x\to 0}{\lim }\log \left(\frac{1}{1+\cot x^2}\right)$

This function can be represented as the composition of 4 functions:

1. $f_1(x)=x^2$
2. $f_2(y)=\cot y$
3. $f_3(z)=(1+z{)}^{-1}$
4. $f_4(u)=\log u$

Since $f_1$ is continuous, we have $\underset{x\to 0}{\lim }{f}_1(x)=f_1(0)=0$. Moreover, $f_1(x)>0$ for $x\ne 0$.

Since $\underset{y\to 0+}{\lim }{f}_2(x)=+\mathrm{\infty }$, it follows that $(f_2\circ f_1)(x)\to +\mathrm{\infty }$ as $x\to 0$.

Further, $f_3(z)\to 0+$ as $z\to +\mathrm{\infty }$ and $f_4(u)\to -\mathrm{\infty }$ as $u\to 0^{+}$.

Finally $(f_4\circ f_3\circ f_2\circ f_1)(x)\to -\mathrm{\infty }$ as $x\to 0$.

Subproblem 2b

$\underset{x\to 64}{\lim }\frac{\sqrt{x}-8}{\sqrt[3]{x}-4}$

This is indeterminate of form $\frac{0}{0}$, but we can do better: Consider function $u(x)=x^{\frac{1}{6}}$ defined on $(0,\mathrm{\infty })$. Since this function is continuous at 64 and $u(64)=2$, we obtain

$\begin{array}{rl}\underset{x\to 64}{\lim }\frac{\sqrt{x}-8}{\sqrt[3]{x}-4}& =\underset{x\to 64}{\lim }\frac{{(u(x))}^3-8}{{(u(x))}^2-4}\\ & =\underset{y\to 2}{\lim }\frac{y^3-8}{y^2-4}\\ & =\underset{y\to 2}{\lim }\frac{(y-2)(y^2+2y+4)}{(y-2)(y+2)}\\ & =\underset{y\to 2}{\lim }\frac{y^2+2y+4}{y+2}\\ & =3\end{array}$

Subproblem 2c

$\underset{n\to \mathrm{\infty }}{\lim }{(1+\frac{c}{n})}^n$, where $c\in \mathbb{ℝ}$.

Let $a_n={(1+\frac{c}{n})}^n$ for $n\in \mathbb{\mathbb{N}}$.

For $n$ large enough, we have $1+\frac{c}{n}>0$, so $a_n>0$. Then

$\log \left({(1+\frac{c}{n})}^n\right)=n\log (1+\frac{c}{n})={\frac{\log (1+cx)}{x}|}_{x=\frac{1}{n}}$

Since $\frac{1}{n}\to 0$ as $n\to \mathrm{\infty }$ and

$\begin{array}{rl}\underset{x\to 0}{\lim }\frac{\log (1+cx)}{x}& ={(\log (1+cx))|}_{x=0}\\ & ={\frac{c}{1+cx}|}_{x=0}\\ & =c\end{array}$

We obtain that ${\log }_{a_n}\to c$ as $n\to \mathrm{\infty }$, therefore $a_n={\mathrm{e}}^{\log a_n}={\mathrm{e}}^c$ as $n\to \mathrm{\infty }$

Problem 3 (20 pts)

Prove that series converges to $\sin x$ for any $x\in \mathbb{ℝ}$:

$\sum _{n=1}^{\mathrm{\infty }}(-1{)}^{n+1}\frac{x^{2n-1}}{(2n-1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}$

Proof. The function $f(x)=\sin x$ is infinitely differentiable on $\mathbb{ℝ}$. According to Taylor's formula, for any $x,x_0\in \mathbb{ℝ}$ and $n\in \mathbb{\mathbb{N}}$,

$f(x)=f(x_0)+\frac{f^{\prime }(x_0)}{1!}(x-x_0)+\dots +\frac{f^{(n)}(x_0)}{n!}(x-x_0{)}^n+R_n(x,x_0)$

where $R_n(x,x_0)=\frac{f^{(n+1)}(\theta )}{(n+1)!}(x-x_0{)}^{n+1}$ for some $\theta =\theta (x,x_0)$ between $x$ and $x_0$. Since $f^{\prime }(x)=\cos x$ and $f^{″}(x)=-\sin x=-f(x)$ for all $x\in \mathbb{ℝ}$, it follows that $f^{(n+1)}(\theta )\le 1$ for all $n\in \mathbb{\mathbb{N}}$ and $\theta \in \mathbb{ℝ}$. Further, one derives that $R_n(x,x_0)\to 0$ as $n\to \mathrm{\infty }$. Thus we obtain an expansion of $\sin x$ into a series.

In the case

$x_0=0$

, this is the required series. (up to zero terms)

quod erat demonstrandum

Problem 4 (20 pts)

Evaluate following integrals

Subproblem 4a

$\int \frac{\sqrt{1+\sqrt[4]{x}}}{2\sqrt{x}}\mathrm{d}x$

To find this integral, we change the variable twice:

1. $\int \frac{\sqrt{1+\sqrt[4]{x}}}{2\sqrt{x}}\mathrm{d}x=\int \sqrt{1+\sqrt{u}}\mathrm{d}u$ for $u=\sqrt{x}$
2. $\int \sqrt{1+\sqrt{u}}\mathrm{d}u=\int (4w^4-4w^2)\mathrm{d}w$ for $w=\sqrt{1+\sqrt{u}}$

The integral of this is $\frac{4}{5}{w}^5-\frac{4}{3}{w}^3+C=\frac{4}{3}{(1+x^{\frac{1}{4}})}^{\frac{3}{2}}+C$.

Subproblem 4b

$\underset{0}{\overset{\sqrt{3}}{\int }}\frac{x^2+6}{x^2+9}\mathrm{d}x$

To evaluate this definite integral, we use linearity of the integral and a substitution $x=3u$:

$\begin{array}{rl}\underset{0}{\overset{\sqrt{3}}{\int }}\frac{x^2+6}{x^2+9}\mathrm{d}x& =\underset{0}{\overset{\sqrt{3}}{\int }}(1-\frac{3}{x^2+9})\mathrm{d}x\\ & =\underset{0}{\overset{\sqrt{3}}{\int }}1\mathrm{d}x-\underset{0}{\overset{\sqrt{3}}{\int }}\frac{3}{x^2+9}\mathrm{d}x\\ & =\sqrt{3}-\underset{0}{\overset{\frac{\sqrt{3}}{3}}{\int }}\frac{3}{(3u{)}^2+9}\mathrm{d}(3u)\\ & =\sqrt{3}-\underset{0}{\overset{\frac{1}{\sqrt{3}}}{\int }}\frac{1}{u^2+1}\mathrm{d}u\\ & =\sqrt{3}-{\arctan u|}_{u=0}^{\frac{1}{\sqrt{3}}A}\\ & =\sqrt{3}-\frac{\pi }{6}-0\end{array}$

Subproblem 4c

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^2{\mathrm{e}}^{-x}\mathrm{d}x$

To evaluate this improper integral, we integrate by parts twice:

$\begin{array}{rl}\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^2{\mathrm{e}}^{-x}\mathrm{d}x& =-\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^2\mathrm{d}\left({\mathrm{e}}^{-x}\right)\\ & ={-x^2{\mathrm{e}}^{-x}|}_0^{\mathrm{\infty }}+\underset{0}{\overset{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-x}\mathrm{d}(x^2)\\ & =\underset{0}{\overset{\mathrm{\infty }}{\int }}2x{\mathrm{e}}^{-x}\mathrm{d}x\\ & ={-2x{\mathrm{e}}^{-x}|}_0^{\mathrm{\infty }}+\underset{0}{\overset{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-x}\mathrm{d}(2x)\\ & =\underset{0}{\overset{\mathrm{\infty }}{\int }}2{\mathrm{e}}^{-x}\mathrm{d}x\\ & ={-2{\mathrm{e}}^{-x}|}_0^{\mathrm{\infty }}\\ & =2\end{array}$

Problem 5 (20 pts)

Check for convergence of series

Subproblem 5a

$\sum _{n=1}^{\mathrm{\infty }}\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}$

This diverges because the sumand terms are just $\frac{1}{{(\sqrt{n+1}+\sqrt{n})}^2}>\frac{1}{4(n+1)}$. The series of the other term diverges because it is the harmonic series, therefore the main series diverges by comparison.

Subproblem 5b

$\sum _{n=1}^{\mathrm{\infty }}\frac{\sqrt{n}+2^n\cos n}{n!}$

This series can be represented as $\sum _{n=1}^{\mathrm{\infty }}(b_n+c_n\cos n)$, where $b_n=\frac{\sqrt{n}}{n!}$ and $c_n=\frac{2^n}{n!}$ for all $n\in \mathbb{\mathbb{N}}$.

The series $\sum _{n=1}^{\mathrm{\infty }}{b}_n$ and $\sum _{n=1}^{\mathrm{\infty }}{c}_n$ both converge due to ratio test, and so does their sum. $|b_n+c_n\cos n|\le b_n+c_n$ for all $n\in \mathbb{\mathbb{N}}$, so the series converges absolutely by the comparison test.

Subproblem 5c

$\sum _{n=2}^{\mathrm{\infty }}\frac{{(-1)}^n}{n\log n}$

This series converges by the alternating series test, but not absolutely due to the integral test.

Bonus Problem 6 (15 pts)

Prove that an infinite product converges:

$\prod _{n=1}^{\mathrm{\infty }}\frac{n^2+1}{n^2}$

Take log of product:

$\log \left(\prod _{n=1}^{\mathrm{\infty }}\frac{n^2+1}{n^2}\right)=\sum _{n=1}^{\mathrm{\infty }}\log \left(\frac{n^2+1}{n^2}\right)$